Wind Pressure. 47* 60 Ibs. per.sq. ft. (according to the exposure) upon the area (k x width of bay). Total force Pt = 56 x k wf, say, in Ibs. Then Pt may be resolved into two forces, one parallel to the rafter, and one (Pn) normally, and Pn total = ~—- ff This force is distributed at a, d, and c as -fV .^n> f Pn, A Pn. In iron roofs the expansion is usually allowed for by fixing one end (a) and leaving the other free (£), which allows us to say the reaction, Rt2, is vertical. Now Rtl, Rt25 Pn are three balancing forces, and must meet in one point X, found by producing Pn to meet Rt2 produced. Then joining a X, the direction of Rtl is found, and the amounts Rtl -and Rt2 further obtained from the auxiliary diagram. If the wind blow from the right, Pn is exerted on cb, and x will be above instead of below b (Case III.). Both II. and III. must be examined, though we only, have space for Case II. Take the lettering as in L, with the exception of the additional external space L, and draw the stress diagrams as shewn below. Finally, tabulate the stresses for Cases L, II., and III.: then add L to II. and III. separately to find the maximum stresses, and design.th'e members to suit. (See Appendix, IV,, p. 953.) Framed Structures of Three Dimensions are such as include a solid instead of an area. They must be solved by a step-by-step process, taking each plane in succession. We will explain by means of an example. Example 39.—A sheer legs (Fig. 433) is formed of two fore legs 145' long and 60' apart at the base, and a back leg- 170' long attached to a nut having a travel of 40'. The maximum overhang is 40', and load too tons. Find the stresses in the members, (i) and (2), at each end of the nut stroke; (3) when the load'is directly over the base plates, (Hons. Mach, Gonstr. Ex.'1888.) Case L Nut at P.—Taking first the plane AD B, the diagram P shews stresses . A _, Turning to the end view, the stress of 174 in AB must be resolved in each leg, as in diagram Q, giving stresses in a b and a c, each =* 88 tons.